# Exponent Challenge Question Series: Day 4

This is the fourth in a series of five blog posts that we will be publishing this week highlighting challenging exponent questions!

This week on Beat The GMAT, Veritas Prep’s authors will each day contribute a difficult exponent-related challenge problem, with a solution to follow the next day. Before you begin, you may want to consider this as a way to crack the exponent code; nearly all exponent-based problems can be solved using a combination of these three guiding exponent principles:

- Exponent rules are almost all related to multiplication and division with virtually no rules that directly apply to addition and subtraction. When facing exponent problems, look for opportunities to factor and multiply to put yourself in a position to use your multiplication-heavy exponent expertise.
- Most exponent rules require you to have common bases in order to apply them, so look to break down bases into prime factors so that you have common bases with which to work.
- Exponents are very pattern-driven, so when large numbers are present you can often establish a rule using small numbers and then extrapolate it to the larger ones.

**Yesterday’s Challenge Question and Solution **

If you haven’t don’t so already, try out yesterday’s problem first before reviewing the solution.

(A)

(B)

(C)

(D)

(E)

C. When working with exponents, you should try to find common (typically prime) bases and multiply. First to multiply, you can factor a out of each additive term within the parentheses to get: . Then, perform the addition within the parentheses to get: .

You can break down to be , which is , and then you have common prime terms for each exponent: .

Multiplying out the equation , you get: and , which can be simplified to .

This can be expressed as , and the correct answer is C.

** ****Today’s Challenge Question**

What is the tens digit of ?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

## 20 comments

saurabh goyal on September 29th, 2010 at 1:31 am

It is happening when ever be multiply a two digit number to any other number we actually multiply in two steps first by ten’s times then by the remaining number of times and then we add then to get the desired result

Like I have to multiply 11 * 11

I don’t know how to write on this but we do something like this 11 in the upper row and 11x in the second row and then we add by this means last term will always be 1. And the second last term will be the sum of tens place of upper term and unit place of lower term.

I really don’t know How much I able to explain

Do this simple way 121 * 11 . ones place term will always be 1 and tens place term will always be the sum of tens place of 121 and tens place of 11 which is equal to 2 +1 = 3

Like this in 11^ 4 == ..31 * 11 ,

tens place will be 3+1=4

Like this in 11^ 5 == ..41 * 11 ,

tens place will be 4+1=5

But for 11^10 == ………..91 * 11

tens place will be 9+1=0

And so on and so for 11^13 === ………………………..21*11

tens place will be 2+1= 3

saurabh goyal on September 29th, 2010 at 1:31 am

Answer is option C

3

ASKI@1st on September 29th, 2010 at 2:13 am

C) – 3

11 is a special number.

Keep in mind that the unit of 11^13 will always be 1 as 1^13=1

Let’s only care about the last two digits of 11^13.

For a number X1 (with X being a digit), we know that the tens of X1*11 is the last digit of Y=X+1, for example the tens of 91*11 is the unit of 9+1=10, by the way 91*11=1001).

So

The tens of 11^2 is equal to 1+1=2

The tens of 11^3 is equal to 2+1=3

……

The tens of 11^10 is equal to 9+1=10 ie 0 as said before

The tens of 11^11 is equal to 0+1=1

….

The tens of 11^13 is equal to 2+1 = 3

I suspect there might be another way to solve this but we’ll come up with the same conclusion.

ASKI

Ankur on September 29th, 2010 at 2:54 am

As mentioned this is exponents question. Is dere any way to solve it by applying laws of exponents?

Gurpreet singh on September 29th, 2010 at 6:20 am

The last two digits can be calculated by calculating the remainder of 11^13 when divided by 100

11* (11^2)^6 = 11* (121)^6

Remainder of 11* (121)^6 = 11* 21^6 = 11* 441^3

Remainder of 11* 441^3 = 11* 41*41*41

Remainder of (11* 41)*(41*41) = 51*81

Remainder of 51*81 = 31

Hence the last two digits are 31 and the answer is C

Sajid Marickar on September 29th, 2010 at 7:35 am

Logic:When ever we mutiply some number by 11, the tens digit of the resulting number = sum of the units digit plus tens digit of that number.

Example: 23*11=253(tens place 5=ones digit 3+ tens digit 2).

Similarly for 11*11=121(ones digit 1=tens digit 1)

If you see, there is a pattern emerging.

11^2=Last 2 digits =21

11^3=Last 2 digits =31

11^4=Last 2 digits =41

11^5=Last 2 digits =51

11^6=Last 2 digits =61

11^7=Last 2 digits =71

11^8=Last 2 digits =81

11^9=Last 2 digits =91

11^10=Last 2 digits =01

11^11=Last 2 digits =11

11^12=Last 2 digits =21

11^13=Last 2 digits =31

(For the logic to click try multiplying some numbers with 11 and try to deduce a pattern)

Tommy Hutomo on September 29th, 2010 at 7:47 am

Somehow exponents of 11 has a same pattern as pascal triangle for the last two numbers.

1 = 11^0

1 1 = 11^1

1 2 1 = 11^2

1 3 3 1 = 11^3

I believe you can now guess what's in 11^13.

............................3 1

Answer is C

Joe H on September 29th, 2010 at 9:34 am

Determine the pattern - the # in the tens digit = the number in the exponent's units digit.

3 --> Answer = C

Scott on September 29th, 2010 at 10:36 am

Answer C, 3.

11, tens digit is 1

11^2=121, tens digit is 2. From there I sensed a potential pattern, but went ahead and did

11^3=1331, tens digit is a 3.

Given 11^13 can be 11^3*11^10, Another ten instances of increasing the tens digit by one would have the 13th instance, and the pattern would repeat.

Yogesh on September 29th, 2010 at 11:02 am

Answer: C

i.e 3

Divya on September 29th, 2010 at 5:21 pm

For all those people who are saying pattern, i get it, however i thought about what if the same question was for 13^13 in which case the pattern would be:

13^1=13 (tens place =1)

13^2=169 (tens place=6)

13^3=2197 (tens place=9)

13^4=28561 (tens place=6)

13^5=371293 (tens place=9)

But if we do 13^6=4826809 (tens place=0) and hence pattern breaks, so we cannot assume there must be a pattern after the first few ones. There has to be a better strategy for solving such questions.

-Divya

Nik on September 29th, 2010 at 10:55 pm

Hi Divya

11 is a special no, that's why the pattern exists. Multiplying by 11 means adding the no with the same no multiplied by 10.

i.e. x* 11 = x * (10+1) = 10x + x

Therefore, 11 ^2 = 11*11 = 110+ 11.

If you observe, the units digit of the nth power of 11 will always be 1. For the tens digit, it will be the tens digit of the (n-1)th power + 1.

e.g. 11 ^ 3 = 121 * 11 = 1210 + 121

Units digit = 0 + 1

Tens = 2 + 1 = 3

This is the theory behind the pattern.

13 just doesn't make the cut !

One more special property for nos with only 1s as digits --

11^2 = 11

111^2 = 12321

1111 ^ 2 = 1234321

11111 ^ 2 = 123454321

Ishaan Singhal on September 30th, 2010 at 4:42 am

Hey gurpreet thanks for the amazing technique to solve the numerical.

It does answer divya's query as well. I tried solving for 13^13 and got the tens and units place correctly. 5 and 3 in tens and units place respectively. takes 30-35 secs to solve it.

Cheerz!!!

Divya on September 30th, 2010 at 4:59 pm

Hey Nik, thanks for your pattern spotting technique. If I follow that, then 11^13= <> * (10+1). So for the tens digit, it will be the tens digit of the 12th power + 1. Now how do i know what the tens digit of the 12th power is? Still no luck

Well i have brooded over this so much that I will probably remm this one, but still cannot find a simpler method to solve such problems...say find the tens digit of 14^24...NO CLUE

Brian Galvin on September 30th, 2010 at 5:51 pm

Hey Divya - good point...and know that they'd never ask you about 14^24 - they're only going to ask you about patterns that you could theoretically find within 2 minutes. 11 is a unique enough number that it has a pretty manageable pattern, so that's why they might ask you for something like this.

Keep in mind - the GMAT is testing your problem solving ability, so they're only going to include problems in which that ability can be put to use. They'll get creative to find awkward situations in which it may not look like it's possible to find an elegant solution, but on ANY GMAT problem there's going to be an efficient way.

Ishaan Singhal on September 30th, 2010 at 6:06 pm

Hey brian thanks for the great question. But like i said before the method used by gurpreet singh can be used to solve this numerical as well. we dont have to look for a pattern always.

For eg:- I will solve divya's problem itself

What is unit's place and Tens place of 14^24

Sol:-

We can get the units place and tens place if we find the remainder when we divide the given number by 100.

Let n = 14^24 = (14^2)^12 = 196^12

Remainder of n when divided by 100 = 96^12 = (96^2)^6 = (xx16)^6.......... we dont need the digits on hundreds place and henceforth

Remainder of (xx16)^6 when divided by 100 = 16^6 = (16^2)^3 = 256^3

Remainder of 256^3 when divided by 100 = 56^3 = (56*56)*56 = (xx36)*56 = (xxxx16)

remainder of xxxx16 when divided by 100 = 16

hence the digit in units place = 6

digit in tens place = 1

Actual value of 14^24 = 3,214,199,700,417,740,936,751,087,616

We can check the same for other questions also.

Ishaan

Divya on September 30th, 2010 at 8:15 pm

Thanks Brian and Ishaan, that was goooood Btw Ishaan tht method is quite good, i tried it for a few numbers and it works. Its good to know, is it called something and can i read abt it somewhere?

Thanks Again!!!

Ishaan Singhal on September 30th, 2010 at 10:52 pm

Hey divya,

I really dunno what do we call the technique. It is just a methodology to find the value at units place. Though i can explain you the idea behind it.

Let us assume i give you a number 23 and ask you what is the tens place in this. You will promptly say that it is 2. Now tell me what is the remainder if i divide 23 by 100. It is 23. So that's how you get the answer. Suppose i ask you what is the digit in hundred's place of 1568. Try finding the remainder when u divide 1568 by 1000. It is 568. So there we have the digit at hundred's place.

Like Brian said, I doubt at GMAT they will ask very tough numbers.

@Brian how about making this technique in a video and sharing with the people. I am sure loads of people will love to know this.

Ishaan

Brian Galvin on October 1st, 2010 at 4:41 pm

Hey Ishaan,

I love it - yeah, let me figure out the best way to share it. Would you mind if I wrote it up as a blog post if I credit you and Gurpreet in the post?

Like you said, the GMAT probably wouldn't test this without a really precise selection of numbers since it does still involve multiple items of 2-digit by 2-digit multiplication which they don't tend to require much, but the theory behind the method is terrific and I know that they've tested 11s before so it's in the realm of fair game.

Thanks!

Brian

Ishaan Singhal on October 1st, 2010 at 4:45 pm

Hey Brian,

Sure, make it as a blog and share with people. It's all about sharing of knowledge, isn't?

Looking forward to the blog.

Thanks,

Ishaan